cutiecomittee123
  • cutiecomittee123
How do i solve this system of equations? y=(x+1)^2-2 and y-2x-3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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whpalmer4
  • whpalmer4
is the second equation \[y=2x-3\]?
whpalmer4
  • whpalmer4
Well, let's assume that it is. \[y = (x+1)^2-2\]\[y=2x-3\] We can solve this by substitution. Everywhere you see \(y\) in the first equation, replace it with the right hand side of the second equation (which is also equal to \(y\)). This gives us an equation all in terms of just the variable \(x\): \[2x-3 = (x+1)^2-2\]Now you can expand that out: \[2x-3 = (x+1)(x+1)-2\]\[2x-3=x^2+1x+1x+1*1-2\]\[2x-3=x^2+2x-1\]and get all the terms on one side and you have a quadratic to solve. You know how to do that?
cutiecomittee123
  • cutiecomittee123
The second equation is y-2x=3

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whpalmer4
  • whpalmer4
Okay, can you solve \[y-2x=3\] for \(y\)?
cutiecomittee123
  • cutiecomittee123
y=3+2x
whpalmer4
  • whpalmer4
Right. So now we do the same thing I did, except with the correct equation. \[y = (x+1)^2+2\]\[y = 2x+3\] \[2x+3 = (x+1)^2 + 2\] Expand both sides, then combine all the terms on one side and solve the quadratic.
cutiecomittee123
  • cutiecomittee123
wait it should be 3+2x=(x+1)^2-2
whpalmer4
  • whpalmer4
doesn't \[2x+3 = 3+2x\]?
whpalmer4
  • whpalmer4
sorry, I switched a sign, that's what you were pointing out, my bad!
whpalmer4
  • whpalmer4
\[3+2x = (x+1)^2-2\]
cutiecomittee123
  • cutiecomittee123
yes but the original equation is y=(x+1)^2-2 and if we are plugging in y , and y=2x+3
cutiecomittee123
  • cutiecomittee123
Yeah thats what I got :))
whpalmer4
  • whpalmer4
Glad one of us is paying close attention :-)
cutiecomittee123
  • cutiecomittee123
so now i get 2x+3=x^2+1+2
cutiecomittee123
  • cutiecomittee123
thats after expanding the equation
whpalmer4
  • whpalmer4
close... \[(x+1)^2 = (x+1)(x+1) = x*x + 1*x + 1*x + 1*1 = x^2 + 2x + 1\] so [2x+3 = x^2+2x+1 - 2\]\[2x+3 = x^2+2x-1\] right?
cutiecomittee123
  • cutiecomittee123
yes got that
whpalmer4
  • whpalmer4
okay, get all the terms collected on one side, what do you get?
cutiecomittee123
  • cutiecomittee123
0=x^2-4
whpalmer4
  • whpalmer4
Okay, and what are the solutions to that equation?
whpalmer4
  • whpalmer4
That's a simple one to solve: \[x^2-4 = 0\]\[x^2 = 4\]Take the square root of both sides and you see \[x=\pm 2\] Have to be sure to include the negative square root or we won't get all of the answers! Now we need to find the values of \(y\) that go along with those values of \(x\). We can use either equation, so might as well use the simpler one: \[y=2x+3\] Plug in both values of \(x\) and get the corresponding values of \(y\).
cutiecomittee123
  • cutiecomittee123
okay just a sec
cutiecomittee123
  • cutiecomittee123
I get (2,7) and (-2,-1)
whpalmer4
  • whpalmer4
So do I!
whpalmer4
  • whpalmer4
and we can easily test them: \[2(2)+3 = (2+1)^2-2\]\[7 = 9-2\checkmark\] \[2(-2)+3 = (-2+1)^2-2\]\[-1=1-2\checkmark\]
whpalmer4
  • whpalmer4
So we can do this any time we can rearrange the stuff in one equation to give us something we can substitute into the other one.
cutiecomittee123
  • cutiecomittee123
sweet thanks
whpalmer4
  • whpalmer4
Here's a tip that doesn't really matter here so much, but does in problems that you will encounter: always try your answers out in the original equation and make sure they work! Whenever you have to square both sides of the equation, you can get what called "spurious" solutions which come out just the same way as the real ones do, but don't actually make the equations true when you plug the numbers in. Rather than remembering exactly what kinds of problems require you to do this, I prefer to simply always check my answers!
anonymous
  • anonymous
Cool! @whpalmer4

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